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Binding a const function reference to a lambda

What is a lambda (function)?What are the differences between a pointer variable and a reference variable in C++?Why are Python lambdas useful?What is the difference between const int*, const int * const, and int const *?How come a non-const reference cannot bind to a temporary object?C++0x lambda capture by value always const?Returning unique_ptr from functionsWhat is a lambda expression in C++11?Returning temporary object and binding to const referenceReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviations

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

16

3

This code

int(&foo)(int, int) = [](int a, int b) return a + b; ;

doesn't compile since apparently a non-const reference can't be initialized with a temporary. Where do I put the const?

c++ c++11 lambda

share|improve this question

edited Jul 1 at 8:21

StoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

asked Jul 1 at 4:33

ArtikashArtikash

31533 silver badges1111 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why does it need to be a const ref? By making a copy the compiler will make use of the move semantics
  
  – Narase
  Jul 1 at 4:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It doesn't compile because a lambda does not have conversion to reference. The compiler error is misleading.
  
  – cpplearner
  Jul 1 at 4:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why not just use auto foo = [](int a, int b) return a + b; ;? C++ will move instead of copy the temporary.
  
  – Yang
  Jul 1 at 4:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It doesn't need to be, I'm just curious how/if this can be done.
  
  – Artikash
  Jul 1 at 4:58
  

  
  
  
  

add a comment | 

16

3

This code

int(&foo)(int, int) = [](int a, int b) return a + b; ;

doesn't compile since apparently a non-const reference can't be initialized with a temporary. Where do I put the const?

c++ c++11 lambda

share|improve this question

edited Jul 1 at 8:21

StoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

asked Jul 1 at 4:33

ArtikashArtikash

31533 silver badges1111 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why does it need to be a const ref? By making a copy the compiler will make use of the move semantics
  
  – Narase
  Jul 1 at 4:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It doesn't compile because a lambda does not have conversion to reference. The compiler error is misleading.
  
  – cpplearner
  Jul 1 at 4:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why not just use auto foo = [](int a, int b) return a + b; ;? C++ will move instead of copy the temporary.
  
  – Yang
  Jul 1 at 4:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It doesn't need to be, I'm just curious how/if this can be done.
  
  – Artikash
  Jul 1 at 4:58
  

  
  
  
  

add a comment | 

This code

int(&foo)(int, int) = [](int a, int b) return a + b; ;

doesn't compile since apparently a non-const reference can't be initialized with a temporary. Where do I put the const?

c++ c++11 lambda

share|improve this question

edited Jul 1 at 8:21

StoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

asked Jul 1 at 4:33

ArtikashArtikash

31533 silver badges1111 bronze badges

int(&foo)(int, int) = [](int a, int b) return a + b; ;

share|improve this question

edited Jul 1 at 8:21

StoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

asked Jul 1 at 4:33

ArtikashArtikash

31533 silver badges1111 bronze badges

share|improve this question

edited Jul 1 at 8:21

StoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

asked Jul 1 at 4:33

ArtikashArtikash

31533 silver badges1111 bronze badges

edited Jul 1 at 8:21

StoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

edited Jul 1 at 8:21

StoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

asked Jul 1 at 4:33

ArtikashArtikash

31533 silver badges1111 bronze badges

asked Jul 1 at 4:33

ArtikashArtikash

31533 silver badges1111 bronze badges

2 Answers
2

active

oldest

votes

24

As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.

int(&foo)(int, int) = *[](int a, int b) return a + b; ;

Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.

Here it is live.

share|improve this answer

answered Jul 1 at 5:26

StoryTellerStoryTeller

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add a comment | 

11

A lambda can only be converted to a function pointer if it does not capture.

The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator

[Lambda Functions][1]

I changed your code as below and it worked.

int (*foo)(int, int)= [] (int a, int b) return a + b; ;
int main()

 cout << "Res:: " << foo(10,20);
 return 0;

I just make it function pointer.

Alternatively,

auto foo = [](int a, int b) return a + b; ; 

is also a good choice.

I hope it helps!

share|improve this answer

answered Jul 1 at 5:06

Abhishek SinhaAbhishek Sinha

38111 silver badge66 bronze badges

add a comment | 

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24

As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.

int(&foo)(int, int) = *[](int a, int b) return a + b; ;

Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.

Here it is live.

share|improve this answer

answered Jul 1 at 5:26

StoryTellerStoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

add a comment | 

24

As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.

int(&foo)(int, int) = *[](int a, int b) return a + b; ;

Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.

Here it is live.

share|improve this answer

answered Jul 1 at 5:26

StoryTellerStoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

add a comment | 

As mentioned already, a capture-less lambda is convertible to a function pointer. So if you want to bind that static function to a reference, you need to dereference the pointer.

int(&foo)(int, int) = *[](int a, int b) return a + b; ;

Applying * to the lambda causes a bunch of machinery to kick in. Since the lambda doesn't overload operator*, but does implement a conversion to a pointer type, that conversion happens. Afterwards * is applied to the returned pointer and that yields a function lvalue. That lvalue can then bind to the reference.

Here it is live.

share|improve this answer

answered Jul 1 at 5:26

StoryTellerStoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

int(&foo)(int, int) = *[](int a, int b) return a + b; ;

share|improve this answer

answered Jul 1 at 5:26

StoryTellerStoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

answered Jul 1 at 5:26

StoryTellerStoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

answered Jul 1 at 5:26

StoryTellerStoryTeller

114k1818 gold badges245245 silver badges311311 bronze badges

11

A lambda can only be converted to a function pointer if it does not capture.

The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator

[Lambda Functions][1]

I changed your code as below and it worked.

int (*foo)(int, int)= [] (int a, int b) return a + b; ;
int main()

 cout << "Res:: " << foo(10,20);
 return 0;

I just make it function pointer.

Alternatively,

auto foo = [](int a, int b) return a + b; ; 

is also a good choice.

I hope it helps!

share|improve this answer

answered Jul 1 at 5:06

Abhishek SinhaAbhishek Sinha

38111 silver badge66 bronze badges

add a comment | 

11

A lambda can only be converted to a function pointer if it does not capture.

The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator

[Lambda Functions][1]

I changed your code as below and it worked.

int (*foo)(int, int)= [] (int a, int b) return a + b; ;
int main()

 cout << "Res:: " << foo(10,20);
 return 0;

I just make it function pointer.

Alternatively,

auto foo = [](int a, int b) return a + b; ; 

is also a good choice.

I hope it helps!

share|improve this answer

answered Jul 1 at 5:06

Abhishek SinhaAbhishek Sinha

38111 silver badge66 bronze badges

add a comment | 

A lambda can only be converted to a function pointer if it does not capture.

The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator

[Lambda Functions][1]

I changed your code as below and it worked.

int (*foo)(int, int)= [] (int a, int b) return a + b; ;
int main()

 cout << "Res:: " << foo(10,20);
 return 0;

I just make it function pointer.

Alternatively,

auto foo = [](int a, int b) return a + b; ; 

is also a good choice.

I hope it helps!

share|improve this answer

answered Jul 1 at 5:06

Abhishek SinhaAbhishek Sinha

38111 silver badge66 bronze badges

[Lambda Functions][1]

int (*foo)(int, int)= [] (int a, int b) return a + b; ;
int main()

 cout << "Res:: " << foo(10,20);
 return 0;

auto foo = [](int a, int b) return a + b; ; 

share|improve this answer

answered Jul 1 at 5:06

Abhishek SinhaAbhishek Sinha

38111 silver badge66 bronze badges

answered Jul 1 at 5:06

Abhishek SinhaAbhishek Sinha

38111 silver badge66 bronze badges

answered Jul 1 at 5:06

Abhishek SinhaAbhishek Sinha

38111 silver badge66 bronze badges